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Exam questions · Maths

Functions, Sequences and Rates of Change

  • 30 exam questions
  • 94 marks
  • 45 quick checks

Functions and Function Notation

Just this lesson
  1. 1 Work out [3 marks]

    The diagram shows a function machine for \(f\). (a) Write down an expression for \(f(x)\). [1 mark] (b) Work out \(f(-2)\). [1 mark] (c) Solve \(f(x) = 15\). [1 mark]

    A function machine with two operations, used to work out the function f of x.
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    Model answer

    (a) \(f(x) = 2x + 7\). (b) \(f(-2) = 2 \times (-2) + 7 = 3\). (c) \(2x + 7 = 15\), so \(x = 4\).

    Mark scheme

    • (a) \(2x + 7\) — B1
    • (b) \(3\) — B1
    • (c) \(4\) — B1
  2. 2 Work out [4 marks]

    \(f(x) = 3x + 2\) and \(g(x) = x - 4\) (a) Work out \(fg(6)\). [2 marks] (b) Show that \(fg(x) = 3x - 10\). [2 marks]

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    Model answer

    (a) \(g(6) = 2\), then \(f(2) = 3 \times 2 + 2 = 8\). (b) \(fg(x) = f(x - 4) = 3(x - 4) + 2 = 3x - 12 + 2 = 3x - 10\).

    Mark scheme

    • (a) \(g(6) = 2\) — M1
    • (a) \(8\) — A1
    • (b) \(3(x - 4) + 2\) — M1
    • (b) \(3x - 12 + 2 = 3x - 10\), with the working shown — A1
  3. 3 Find [3 marks]

    \(f(x) = 5x - 2\) (a) Find \(f^{-1}(x)\). [2 marks] (b) Work out \(f^{-1}(8)\). [1 mark]

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    Model answer

    (a) \(y = 5x - 2\), so \(x = \dfrac{y + 2}{5}\). So \(f^{-1}(x) = \dfrac{x + 2}{5}\). (b) \(f^{-1}(8) = \dfrac{10}{5} = 2\).

    Mark scheme

    • (a) \(5x = y + 2\) or equivalent — M1
    • (a) \(f^{-1}(x) = \dfrac{x + 2}{5}\) — A1
    • (b) \(2\) — B1
  4. 4 Solve [4 marks]

    \(f(x) = 3x - 1\) (a) Find \(f^{-1}(x)\). [2 marks] (b) Solve \(f^{-1}(x) = f(x)\). [2 marks]

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    Model answer

    (a) \(f^{-1}(x) = \dfrac{x + 1}{3}\). (b) \(\dfrac{x + 1}{3} = 3x - 1\), so \(x + 1 = 9x - 3\), \(4 = 8x\) and \(x = \dfrac{1}{2}\).

    Mark scheme

    • (a) \(x = \dfrac{y + 1}{3}\) or equivalent — M1
    • (a) \(f^{-1}(x) = \dfrac{x + 1}{3}\) — A1
    • (b) \(\dfrac{x + 1}{3} = 3x - 1\) — M1
    • (b) \(\dfrac{1}{2}\) — A1
  5. 5 Find [4 marks]

    \(f(x) = \dfrac{x + 1}{x - 2}\) where \(x \ne 2\). Find \(f^{-1}(x)\). [4 marks]

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    Model answer

    \(y = \dfrac{x + 1}{x - 2}\), so \(y(x - 2) = x + 1\) and \(xy - 2y = x + 1\). Then \(xy - x = 2y + 1\), so \(x(y - 1) = 2y + 1\) and \(x = \dfrac{2y + 1}{y - 1}\). So \(f^{-1}(x) = \dfrac{2x + 1}{x - 1}\).

    Mark scheme

    • \(y(x - 2) = x + 1\) — M1
    • \(xy - 2y = x + 1\) — M1
    • \(x(y - 1) = 2y + 1\) — M1
    • \(f^{-1}(x) = \dfrac{2x + 1}{x - 1}\) — A1
  6. 6 Work out [4 marks]

    \(f(x) = 2x + 1\) and \(g(x) = ax - 3\), where \(a\) is a constant. \(fg(x) = gf(x)\). Work out the value of \(a\). [4 marks]

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    Model answer

    \(fg(x) = 2(ax - 3) + 1 = 2ax - 5\) and \(gf(x) = a(2x + 1) - 3 = 2ax + a - 3\). So \(-5 = a - 3\), which gives \(a = -2\).

    Mark scheme

    • \(fg(x) = 2ax - 5\) — M1
    • \(gf(x) = 2ax + a - 3\) — M1
    • \(-5 = a - 3\) — M1
    • \(-2\) — A1

Quick check

  1. 1

    \(f(x) = 3x - 5\). What is \(f(4)\)?

    1. A12
    2. B7
    3. C2
    4. D-5
    Show answerHide answer

    B: 7

    \(3 \times 4 - 5 = 7\).

  2. 2

    \(f(x) = 2x + 1\). What is \(f(-3)\)?

    1. A\(-5\)
    2. B7
    3. C\(-6\)
    4. D5
    Show answerHide answer

    A: \(-5\)

    \(2 \times (-3) + 1 = -5\).

  3. 3

    \(f(x) = x^2 + 1\) and \(g(x) = 2x\). What is \(fg(2)\)?

    1. A10
    2. B8
    3. C5
    4. D17
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    D: 17

    \(g(2) = 4\), then \(f(4) = 16 + 1 = 17\).

  4. 4

    \(f(x) = x + 3\) and \(g(x) = x^2\). What is \(gf(x)\)?

    1. A\(x^2 + 3\)
    2. B\(x^2 + 9\)
    3. C\((x + 3)^2\)
    4. D\(x + 9\)
    Show answerHide answer

    C: \((x + 3)^2\)

    \(gf(x) = g(f(x)) = (x + 3)^2\).

  5. 5

    \(f(x) = 2x + 1\). What is \(ff(x)\)?

    1. A\(4x + 1\)
    2. B\(4x + 3\)
    3. C\(4x^2 + 1\)
    4. D\(2x + 2\)
    Show answerHide answer

    B: \(4x + 3\)

    \(ff(x) = 2(2x + 1) + 1 = 4x + 3\).

  6. 6

    What is the inverse of \(f(x) = x + 7\)?

    1. A\(f^{-1}(x) = x - 7\)
    2. B\(f^{-1}(x) = \dfrac{1}{x + 7}\)
    3. C\(f^{-1}(x) = 7 - x\)
    4. D\(f^{-1}(x) = x + 7\)
    Show answerHide answer

    A: \(f^{-1}(x) = x - 7\)

    The inverse reverses the rule, so you subtract 7.

  7. 7

    What is the inverse of \(f(x) = 3x - 5\)?

    1. A\(\dfrac{x - 5}{3}\)
    2. B\(\dfrac{x}{3} + 5\)
    3. C\(\dfrac{1}{3x - 5}\)
    4. D\(\dfrac{x + 5}{3}\)
    Show answerHide answer

    D: \(\dfrac{x + 5}{3}\)

    \(y = 3x - 5\) gives \(x = \dfrac{y + 5}{3}\).

  8. 8

    \(f(x) = 5x - 4\). Solve \(f^{-1}(x) = f(x)\).

    1. A\(x = 0\)
    2. B\(x = -1\)
    3. C\(x = 1\)
    4. D\(x = 2\)
    Show answerHide answer

    C: \(x = 1\)

    \(f^{-1}(x) = \dfrac{x + 4}{5}\), so \(\dfrac{x + 4}{5} = 5x - 4\), which gives \(24x = 24\).

  9. 9

    \(f(x) = 2x - 1\) and \(g(x) = ax + 3\), and \(fg(x) = gf(x)\). What is \(a\)?

    1. A\(a = 2\)
    2. B\(a = -2\)
    3. C\(a = 8\)
    4. D\(a = -8\)
    Show answerHide answer

    B: \(a = -2\)

    \(fg(x) = 2(ax + 3) - 1 = 2ax + 5\) and \(gf(x) = a(2x - 1) + 3 = 2ax - a + 3\), so \(5 = 3 - a\) and \(a = -2\).

Geometric and Special Sequences

Just this lesson
  1. 1 Work out [3 marks]

    Here are the first four terms of a geometric sequence: \(3, 12, 48, 192\) (a) Write down the common ratio. [1 mark] (b) Work out the next two terms. [2 marks]

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    Model answer

    (a) \(12 \div 3 = 4\). (b) \(192 \times 4 = 768\) and \(768 \times 4 = 3072\).

    Mark scheme

    • (a) \(4\) — B1
    • (b) \(768\) — B1
    • (b) \(3072\) — B1
  2. 2 Work out [2 marks]

    The first two terms of a sequence are 2 and 7. Each term after that is the sum of the two terms before it. Work out the 6th term. [2 marks]

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    Model answer

    The terms are \(2, 7, 9, 16, 25, 41\), so the 6th term is 41.

    Mark scheme

    • Continues the sequence, \(9, 16, 25\) — M1
    • \(41\) — A1
  3. 3 Work out [2 marks]

    The first term of a geometric sequence is 81 and the common ratio is \(\dfrac{1}{3}\). Work out the 4th term. [2 marks]

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    Model answer

    \(81, 27, 9, 3\), so the 4th term is 3.

    Mark scheme

    • \(81 \times \left(\dfrac{1}{3}\right)^3\) or \(81, 27, 9\) — M1
    • \(3\) — A1
  4. 4 Work out [3 marks]

    The 2nd term of a geometric sequence is 12 and the 5th term is 324. Work out the first term. [3 marks]

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    Model answer

    There are 3 steps from the 2nd to the 5th term, so \(r^3 = \dfrac{324}{12} = 27\) and \(r = 3\). The first term is \(12 \div 3 = 4\).

    Mark scheme

    • \(r^3 = \dfrac{324}{12} = 27\) — M1
    • \(r = 3\) — A1
    • \(4\) — A1
  5. 5 Work out [3 marks]

    The first four terms of a geometric sequence are \(4, 4\sqrt{2}, 8, 8\sqrt{2}\). Work out the 6th term. [3 marks]

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    Model answer

    The common ratio is \(\sqrt{2}\). The 5th term is \(8\sqrt{2} \times \sqrt{2} = 16\), and the 6th term is \(16\sqrt{2}\).

    Mark scheme

    • Common ratio \(\sqrt{2}\) — B1
    • \(8\sqrt{2} \times \sqrt{2} = 16\) for the 5th term — M1
    • \(16\sqrt{2}\) — A1
  6. 6 Work out [3 marks]

    \(2, x, 18\) are three consecutive terms of a geometric sequence. All the terms are positive. Work out the value of \(x\). [3 marks]

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    Model answer

    \(\dfrac{x}{2} = \dfrac{18}{x}\), so \(x^2 = 36\) and \(x = 6\).

    Mark scheme

    • \(\dfrac{x}{2} = \dfrac{18}{x}\) — M1
    • \(x^2 = 36\) — M1
    • \(6\) — A1

Quick check

  1. 1

    What is the common ratio of \(3, 12, 48, 192\)?

    1. A9
    2. B3
    3. C4
    4. D36
    Show answerHide answer

    C: 4

    \(12 \div 3 = 4\).

  2. 2

    What is the next term of \(2, 6, 18, 54\)?

    1. A108
    2. B162
    3. C72
    4. D216
    Show answerHide answer

    B: 162

    Multiply by 3: \(54 \times 3 = 162\).

  3. 3

    What is the next term in the Fibonacci-type sequence \(3, 5, 8, 13\)?

    1. A21
    2. B18
    3. C26
    4. D16
    Show answerHide answer

    A: 21

    \(8 + 13 = 21\).

  4. 4

    What is the common ratio of \(80, 40, 20, 10\)?

    1. A2
    2. B\(-2\)
    3. C\(\dfrac{1}{4}\)
    4. D\(\dfrac{1}{2}\)
    Show answerHide answer

    D: \(\dfrac{1}{2}\)

    \(40 \div 80 = \dfrac{1}{2}\).

  5. 5

    What is the 5th term of the geometric sequence \(2, 6, 18, \ldots\)?

    1. A54
    2. B486
    3. C162
    4. D90
    Show answerHide answer

    C: 162

    \(2 \times 3^4 = 162\).

  6. 6

    Which of these sequences is geometric?

    1. A\(1, 3, 5, 7\)
    2. B\(1, 3, 9, 27\)
    3. C\(1, 4, 9, 16\)
    4. D\(1, 1, 2, 3\)
    Show answerHide answer

    B: \(1, 3, 9, 27\)

    Each term is multiplied by 3.

  7. 7

    The 2nd term of a geometric sequence is 6 and the 5th term is 48. What is the common ratio?

    1. A2
    2. B3
    3. C8
    4. D4
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    A: 2

    \(r^3 = \dfrac{48}{6} = 8\), so \(r = 2\).

  8. 8

    What is the common ratio of \(2, 2\sqrt{3}, 6, 6\sqrt{3}\)?

    1. A3
    2. B\(2\sqrt{3}\)
    3. C\(\dfrac{1}{\sqrt{3}}\)
    4. D\(\sqrt{3}\)
    Show answerHide answer

    D: \(\sqrt{3}\)

    \(\dfrac{2\sqrt{3}}{2} = \sqrt{3}\).

  9. 9

    A geometric sequence has first term 3 and common ratio 2. What is the \(n\)th term?

    1. A\(3 \times 2^n\)
    2. B\(3n + 2\)
    3. C\(3 \times 2^{n-1}\)
    4. D\(2 \times 3^{n-1}\)
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    C: \(3 \times 2^{n-1}\)

    The \(n\)th term is \(ar^{n-1}\).

Iteration

Just this lesson
  1. 1 Work out [2 marks]

    \(x_{n+1} = 5 - \dfrac{6}{x_n}\) and \(x_0 = 6\). Work out \(x_1\) and \(x_2\). [2 marks]

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    Model answer

    \(x_1 = 5 - \dfrac{6}{6} = 4\) and \(x_2 = 5 - \dfrac{6}{4} = 3.5\).

    Mark scheme

    • \(x_1 = 4\) — B1
    • \(x_2 = 3.5\) — B1
  2. 2 Show that [2 marks]

    Show that the equation \(x^2 - 5x + 6 = 0\) can be rearranged to give \(x = 5 - \dfrac{6}{x}\). [2 marks]

    Show answerHide answer

    Model answer

    Divide every term by \(x\): \(x - 5 + \dfrac{6}{x} = 0\). Then \(x = 5 - \dfrac{6}{x}\).

    Mark scheme

    • Divides by \(x\), giving \(x - 5 + \dfrac{6}{x} = 0\) — M1
    • \(x = 5 - \dfrac{6}{x}\) — Q1
  3. 3 Work out [4 marks]

    \(x_{n+1} = 1 + \dfrac{6}{x_n}\). The values \(x_n\) tend to a positive limit \(a\). Show that \(a^2 - a - 6 = 0\), and work out the value of \(a\). [4 marks]

    Show answerHide answer

    Model answer

    At the limit, \(a = 1 + \dfrac{6}{a}\), so \(a^2 = a + 6\) and \(a^2 - a - 6 = 0\). Then \((a - 3)(a + 2) = 0\), and \(a\) is positive, so \(a = 3\).

    Mark scheme

    • \(a = 1 + \dfrac{6}{a}\) — M1
    • \(a^2 - a - 6 = 0\) — M1
    • \((a - 3)(a + 2) = 0\) — M1
    • \(3\) — A1
  4. 4 Show that [4 marks]

    \(f(x) = x^3 + 2x - 5\) (a) Show that the equation \(f(x) = 0\) has a root between 1 and 2. [2 marks] (b) Work out this root to 1 decimal place. Show your working. [2 marks]

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    Model answer

    (a) \(f(1) = -2\) and \(f(2) = 7\). The sign changes, so there is a root between 1 and 2. (b) \(f(1.3) = -0.203\) and \(f(1.4) = 0.544\), and \(f(1.35) = 0.160\ldots\), which is positive, so the root is between 1.3 and 1.35, which is 1.3 to 1 decimal place.

    Mark scheme

    • (a) \(f(1) = -2\) and \(f(2) = 7\) — M1
    • (a) A change of sign, so there is a root — Q1
    • (b) \(f(1.3) < 0\), \(f(1.4) > 0\) and a test of 1.35 — M1
    • (b) \(1.3\) — A1
  5. 5 Work out [4 marks]

    Use trial and improvement to find the value of \(\sqrt{13}\) correct to 1 decimal place. Show all your working. [4 marks]

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    Model answer

    \(3.6^2 = 12.96\) and \(3.7^2 = 13.69\), so \(\sqrt{13}\) is between 3.6 and 3.7. \(3.65^2 = 13.3225\), which is more than 13, so \(\sqrt{13}\) is below 3.65. So \(\sqrt{13} = 3.6\) to 1 decimal place.

    Mark scheme

    • \(3.6^2 = 12.96\) and \(3.7^2 = 13.69\) — B1
    • Tests \(3.65\) — M1
    • \(3.65^2 = 13.3225\) — A1
    • \(3.6\) — A1
  6. 6 Explain [2 marks]

    \(x_{n+1} = 5 - \dfrac{6}{x_n}\). Explain what happens to the values of \(x_n\) if \(x_0 = 3\). [2 marks]

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    Model answer

    \(x_1 = 5 - \dfrac{6}{3} = 3\), so every term is 3. The sequence stays at 3.

    Mark scheme

    • \(x_1 = 5 - 2 = 3\) — M1
    • States that every term is 3, because the value does not change — Q1

Quick check

  1. 1

    \(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_0 = 4\). What is \(x_1\)?

    1. A3.5
    2. B2
    3. C0.5
    4. D2.5
    Show answerHide answer

    D: 2.5

    \(3 - \dfrac{2}{4} = 3 - 0.5 = 2.5\).

  2. 2

    What does \(x_0\) mean in an iteration?

    1. AThe limit
    2. BThe last value
    3. CThe starting value
    4. DThe number of steps
    Show answerHide answer

    C: The starting value

    \(x_0\) is where the iteration starts.

  3. 3

    At the limit of an iteration, which statement is true?

    1. A\(x_{n+1} = 0\)
    2. B\(x_{n+1} = x_n\)
    3. C\(x_n = 1\)
    4. D\(x_{n+1} = 2x_n\)
    Show answerHide answer

    B: \(x_{n+1} = x_n\)

    At the limit the values stop changing.

  4. 4

    \(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_1 = 2.5\). What is \(x_2\)?

    1. A2.2
    2. B2.8
    3. C2.4
    4. D1.8
    Show answerHide answer

    A: 2.2

    \(3 - \dfrac{2}{2.5} = 3 - 0.8 = 2.2\).

  5. 5

    Which equation can be rearranged to \(x = 3 - \dfrac{2}{x}\)?

    1. A\(x^2 + 3x + 2 = 0\)
    2. B\(x^2 - 2x + 3 = 0\)
    3. C\(x^2 - 3x - 2 = 0\)
    4. D\(x^2 - 3x + 2 = 0\)
    Show answerHide answer

    D: \(x^2 - 3x + 2 = 0\)

    Divide \(x^2 - 3x + 2 = 0\) by \(x\) to get \(x - 3 + \dfrac{2}{x} = 0\).

  6. 6

    \(f(x) = x^3 + x - 3\). What does \(f(1) = -1\) and \(f(2) = 7\) show?

    1. AThe root is 1.5
    2. BThere is no root
    3. CA root lies between 1 and 2
    4. DThe root is 0
    Show answerHide answer

    C: A root lies between 1 and 2

    The sign changes, so a root lies between them.

  7. 7

    \(x_{n+1} = 1 + \dfrac{6}{x_n}\) converges to a positive number \(a\). Which equation does \(a\) satisfy?

    1. A\(a^2 + a - 6 = 0\)
    2. B\(a^2 - a - 6 = 0\)
    3. C\(a^2 - 6a - 1 = 0\)
    4. D\(a^2 - a + 6 = 0\)
    Show answerHide answer

    B: \(a^2 - a - 6 = 0\)

    \(a = 1 + \dfrac{6}{a}\), so \(a^2 = a + 6\).

  8. 8

    What is the value of \(a\) in the previous question?

    1. A3
    2. B\(-2\)
    3. C6
    4. D2
    Show answerHide answer

    A: 3

    \((a - 3)(a + 2) = 0\) and \(a\) is positive.

  9. 9

    \(f(x) = x^2 - 7\). \(f(2.6) = -0.24\) and \(f(2.7) = 0.29\). Why is the root 2.6 to 1 decimal place, given \(f(2.65) = 0.0225\)?

    1. AThe root is 2.7 because \(f(2.7)\) is positive
    2. BThe root is exactly 2.65
    3. CIt cannot be rounded
    4. DThe root is below 2.65, so it rounds down to 2.6
    Show answerHide answer

    D: The root is below 2.65, so it rounds down to 2.6

    \(f(2.65) > 0\) means the root is between 2.6 and 2.65.

Rates of Change and Areas Under Graphs

Just this lesson
  1. 1 Work out [3 marks]

    The diagram shows the graph of \(y = x^2\) and the tangent to the curve at the point \((3, 9)\). Work out the gradient of the curve at the point \((3, 9)\). [3 marks]

    A curve with a tangent drawn at a marked point, on a grid.
    Show answerHide answer

    Model answer

    The tangent passes through \((2, 3)\) and \((4, 15)\). Gradient \(= \dfrac{15 - 3}{4 - 2} = 6\).

    Mark scheme

    • Two points read from the tangent, such as \((2, 3)\) and \((4, 15)\) — M1
    • \(\dfrac{15 - 3}{4 - 2}\) — M1
    • \(6\) — A1
  2. 2 Work out [3 marks]

    The graph shows the distance, \(s\) metres, travelled by a cyclist after \(t\) seconds. The line is the tangent to the curve at \(t = 4\). Work out the speed of the cyclist at \(t = 4\). [3 marks]

    A distance-time graph with a tangent drawn at a marked point.
    Show answerHide answer

    Model answer

    The tangent passes through \((2, 0)\) and \((6, 16)\). Gradient \(= \dfrac{16}{4} = 4\), so the speed is 4 m/s.

    Mark scheme

    • Two points read from the tangent, such as \((2, 0)\) and \((6, 16)\) — M1
    • \(\dfrac{16 - 0}{6 - 2}\) — M1
    • \(4\) m/s — A1
  3. 3 Work out [2 marks]

    The distance, \(s\) metres, travelled by a car after \(t\) seconds is given by \(s = 2t^2\). Work out the average speed of the car between \(t = 1\) and \(t = 3\). [2 marks]

    Show answerHide answer

    Model answer

    \(s = 2\) when \(t = 1\) and \(s = 18\) when \(t = 3\). Average speed \(= \dfrac{18 - 2}{3 - 1} = 8\) m/s.

    Mark scheme

    • \(\dfrac{18 - 2}{3 - 1}\) — M1
    • \(8\) m/s — A1
  4. 4 Work out [5 marks]

    The graph shows the velocity, \(v\) m/s, of a particle at time \(t\) seconds. (a) Use 3 strips of equal width to estimate the distance travelled between \(t = 0\) and \(t = 6\). [3 marks] (b) Is your answer an underestimate or an overestimate? Give a reason for your answer. [2 marks]

    A velocity-time curve with vertical lines at equal intervals, used to estimate the area under it.
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    Model answer

    (a) The heights are 0, 8, 8 and 0. Area \(= \dfrac{1}{2} \times 2 \times (0 + 8) + \dfrac{1}{2} \times 2 \times (8 + 8) + \dfrac{1}{2} \times 2 \times (8 + 0) = 8 + 16 + 8 = 32\) m. (b) An underestimate, because the curve bends downwards and the straight tops of the trapezia are below the curve.

    Mark scheme

    • (a) Reads the heights 8 and 8 — B1
    • (a) Uses \(\dfrac{1}{2}(a + b)h\) for each strip — M1
    • (a) \(32\) — A1
    • (b) Underestimate — B1
    • (b) The curve is above the straight tops of the trapezia — Q1
  5. 5 Work out [3 marks]

    The graph shows the velocity, \(v\) m/s, of a car at time \(t\) seconds. The line is the tangent to the curve at \(t = 2\). Work out an estimate of the acceleration of the car at \(t = 2\). [3 marks]

    A velocity-time graph with a tangent drawn at a marked point.
    Show answerHide answer

    Model answer

    The tangent passes through \((1, 0)\) and \((3, 8)\). Gradient \(= \dfrac{8}{2} = 4\), so the acceleration is 4 m/s\(^2\).

    Mark scheme

    • Two points read from the tangent, such as \((1, 0)\) and \((3, 8)\) — M1
    • \(\dfrac{8 - 0}{3 - 1}\) — M1
    • \(4\) m/s\(^2\) — A1
  6. 6 Work out [4 marks]

    The velocity of a particle was measured every second. \(t\) (s): 0, 1, 2, 3, 4 \(v\) (m/s): 0, 2, 6, 12, 20 Use trapezia to estimate the distance travelled in the first 4 seconds. State whether your answer is an underestimate or an overestimate. Give a reason. [4 marks]

    Show answerHide answer

    Model answer

    Area \(= \dfrac{1}{2}(0 + 2) + \dfrac{1}{2}(2 + 6) + \dfrac{1}{2}(6 + 12) + \dfrac{1}{2}(12 + 20) = 1 + 4 + 9 + 16 = 30\) m. It is an overestimate, because the velocity curve bends upwards, so the straight tops of the trapezia are above the curve.

    Mark scheme

    • \(\dfrac{1}{2}(0 + 2) + \dfrac{1}{2}(2 + 6) + \ldots\), with strips of width 1 — M1
    • \(30\) — A1
    • Overestimate — B1
    • The curve bends upwards, so the straight tops are above the curve — Q1

Quick check

  1. 1

    What do you draw to find the gradient of a curve at a point?

    1. AA tangent
    2. BA chord
    3. CA normal
    4. DA diameter
    Show answerHide answer

    A: A tangent

    A tangent touches the curve at that point.

  2. 2

    What is a chord?

    1. AA line that touches a curve at one point
    2. BA line through the origin
    3. CThe highest point of a curve
    4. DA straight line joining two points on a curve
    Show answerHide answer

    D: A straight line joining two points on a curve

    The chord joins two points on the curve.

  3. 3

    What does the gradient of a distance-time graph represent?

    1. AAcceleration
    2. BDistance
    3. CSpeed
    4. DTime
    Show answerHide answer

    C: Speed

    Distance divided by time is speed.

  4. 4

    What does the area under a velocity-time graph represent?

    1. ASpeed
    2. BDistance travelled
    3. CAcceleration
    4. DTime
    Show answerHide answer

    B: Distance travelled

    Velocity multiplied by time is distance.

  5. 5

    A tangent passes through \((1, 0)\) and \((3, 8)\). What is its gradient?

    1. A4
    2. B8
    3. C2
    4. D\(\dfrac{1}{4}\)
    Show answerHide answer

    A: 4

    \(\dfrac{8 - 0}{3 - 1} = 4\).

  6. 6

    What is the average rate of change of \(y = x^2\) between \(x = 1\) and \(x = 4\)?

    1. A3
    2. B15
    3. C4
    4. D5
    Show answerHide answer

    D: 5

    \(\dfrac{16 - 1}{4 - 1} = 5\).

  7. 7

    What is the area of a trapezium with parallel sides 4 and 6 and width 2?

    1. A20
    2. B5
    3. C10
    4. D24
    Show answerHide answer

    C: 10

    \(\dfrac{1}{2}(4 + 6) \times 2 = 10\).

  8. 8

    A velocity-time curve bends downwards. Is a trapezium estimate of the area an over- or underestimate?

    1. AAn overestimate
    2. BAn underestimate
    3. CIt is exact
    4. DIt depends on the width
    Show answerHide answer

    B: An underestimate

    The straight tops lie below the curve.

  9. 9

    A velocity-time graph has heights 0, 8, 8, 0 at times 0, 2, 4, 6. What is the trapezium estimate of the distance?

    1. A32
    2. B16
    3. C48
    4. D24
    Show answerHide answer

    A: 32

    \(8 + 16 + 8 = 32\).

Transformations of Graphs

Just this lesson
  1. 1 Write down [2 marks]

    The graph of \(y = f(x)\) is shown. The turning point is \((3, -4)\). (a) Write down the coordinates of the turning point of the graph of \(y = f(x) - 2\). [1 mark] (b) Write down the coordinates of the turning point of the graph of \(y = f(x - 2)\). [1 mark]

    The graph of y equals f of x with its turning point marked.
    Show answerHide answer

    Model answer

    (a) \((3, -6)\). (b) \((5, -4)\).

    Mark scheme

    • (a) \((3, -6)\) — B1
    • (b) \((5, -4)\) — B1
  2. 2 Write down [2 marks]

    The graph of \(y = g(x)\) has a minimum point at \((-2, -5)\). (a) Write down the coordinates of the maximum point of \(y = -g(x)\). [1 mark] (b) Write down the coordinates of the minimum point of \(y = g(-x)\). [1 mark]

    Show answerHide answer

    Model answer

    (a) \((-2, 5)\). (b) \((2, -5)\).

    Mark scheme

    • (a) \((-2, 5)\) — B1
    • (b) \((2, -5)\) — B1
  3. 3 Write down [3 marks]

    The graph of \(y = x^2\) is transformed. Write down the equation of the new graph after (a) a translation of 4 units to the left, [1 mark] (b) a translation of 3 units down, [1 mark] (c) a translation by the vector \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\). [1 mark]

    Show answerHide answer

    Model answer

    (a) \(y = (x + 4)^2\). (b) \(y = x^2 - 3\). (c) \(y = (x - 2)^2 - 3\).

    Mark scheme

    • (a) \(y = (x + 4)^2\) — B1
    • (b) \(y = x^2 - 3\) — B1
    • (c) \(y = (x - 2)^2 - 3\) — B1
  4. 4 Write down [4 marks]

    The graph of \(y = f(x)\) has a maximum point at \((-2, 6)\). Write down the coordinates of the maximum point of the graph of (a) \(y = f(x + 3)\) [1 mark] (b) \(y = f(x) + 1\) [1 mark] (c) \(y = -f(x)\) [1 mark] (d) \(y = f(-x)\) [1 mark]

    Show answerHide answer

    Model answer

    (a) \((-5, 6)\). (b) \((-2, 7)\). (c) \((-2, -6)\), which is now a minimum. (d) \((2, 6)\).

    Mark scheme

    • (a) \((-5, 6)\) — B1
    • (b) \((-2, 7)\) — B1
    • (c) \((-2, -6)\) — B1
    • (d) \((2, 6)\) — B1
  5. 5 Write down [3 marks]

    The diagram shows the graph of \(y = \sin x\) and a transformation of it, for \(0^\circ \le x \le 360^\circ\). (a) Write down the equation of the transformed graph. [1 mark] (b) Describe fully the single transformation. [2 marks]

    The graph of y equals sin x and a transformed curve, a sine curve moved 90 degrees to the right.
    Show answerHide answer

    Model answer

    (a) \(y = \sin(x - 90^\circ)\). (b) A translation by the vector \(\begin{pmatrix} 90 \\ 0 \end{pmatrix}\), which is \(90^\circ\) to the right.

    Mark scheme

    • (a) \(y = \sin(x - 90^\circ)\) — B1
    • (b) Translation — B1
    • (b) Vector \(\begin{pmatrix} 90 \\ 0 \end{pmatrix}\), or \(90^\circ\) to the right — B1
  6. 6 Work out [4 marks]

    \(f(x) = x^2 - 6x + 5\). Solve \(f(x - 2) = 0\). [4 marks]

    Show answerHide answer

    Model answer

    \(f(x) = (x - 1)(x - 5)\), so \(f(x - 2) = (x - 3)(x - 7)\). So \(x = 3\) or \(x = 7\).

    Mark scheme

    • \(f(x) = (x - 1)(x - 5)\) — B1
    • \(f(x - 2) = (x - 2 - 1)(x - 2 - 5)\) or \((x - 3)(x - 7)\) — M1
    • \(x = 3\) — A1
    • \(x = 7\) — A1

Quick check

  1. 1

    What does \(y = f(x) + 3\) do to the graph of \(y = f(x)\)?

    1. AMoves it right 3
    2. BMoves it up 3
    3. CMoves it left 3
    4. DMoves it down 3
    Show answerHide answer

    B: Moves it up 3

    Adding to the function moves the graph up.

  2. 2

    What does \(y = f(x + 2)\) do to the graph of \(y = f(x)\)?

    1. AMoves it left 2
    2. BMoves it right 2
    3. CMoves it up 2
    4. DMoves it down 2
    Show answerHide answer

    A: Moves it left 2

    A plus inside the bracket moves the graph left.

  3. 3

    What does \(y = -f(x)\) do to the graph of \(y = f(x)\)?

    1. AReflects it in the \(y\)-axis
    2. BMoves it down
    3. CTurns it by \(180^\circ\) about the origin
    4. DReflects it in the \(x\)-axis
    Show answerHide answer

    D: Reflects it in the \(x\)-axis

    A minus outside changes the \(y\)-values.

  4. 4

    What does \(y = f(-x)\) do to the graph of \(y = f(x)\)?

    1. AReflects it in the \(x\)-axis
    2. BMoves it left
    3. CReflects it in the \(y\)-axis
    4. DMoves it down
    Show answerHide answer

    C: Reflects it in the \(y\)-axis

    A minus inside changes the \(x\)-values.

  5. 5

    The maximum of \(y = f(x)\) is at \((3, 5)\). Where is the maximum of \(y = f(x - 2)\)?

    1. A\((1, 5)\)
    2. B\((5, 5)\)
    3. C\((3, 7)\)
    4. D\((3, 3)\)
    Show answerHide answer

    B: \((5, 5)\)

    The graph moves right 2.

  6. 6

    The maximum of \(y = f(x)\) is at \((3, 5)\). Where is the turning point of \(y = -f(x)\)?

    1. A\((3, -5)\)
    2. B\((-3, 5)\)
    3. C\((-3, -5)\)
    4. D\((3, 5)\)
    Show answerHide answer

    A: \((3, -5)\)

    The \(y\)-coordinate changes sign.

  7. 7

    What is the equation of \(y = x^2\) after a translation of 3 units to the right?

    1. A\(y = (x + 3)^2\)
    2. B\(y = x^2 + 3\)
    3. C\(y = x^2 - 3\)
    4. D\(y = (x - 3)^2\)
    Show answerHide answer

    D: \(y = (x - 3)^2\)

    Moving right replaces \(x\) with \(x - 3\).

  8. 8

    What is the maximum value of \(y = \sin x + 1\)?

    1. A1
    2. B0
    3. C2
    4. D\(-1\)
    Show answerHide answer

    C: 2

    The sine graph is moved up by 1, so its maximum is \(1 + 1 = 2\).

  9. 9

    Which equation gives the same graph as \(y = \cos x\)?

    1. A\(y = \sin(x - 90^\circ)\)
    2. B\(y = \sin(x + 90^\circ)\)
    3. C\(y = -\sin x\)
    4. D\(y = \sin x + 90\)
    Show answerHide answer

    B: \(y = \sin(x + 90^\circ)\)

    Moving the sine graph left by \(90^\circ\) gives the cosine graph.